Answer:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Explanation:
Hello!
1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?
- <u><em>We have the following data:</em></u>
Vo (initial volume) = 1.00 L
V (final volume) = 473 mL → 0.473 L
Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)
P (final pressure) = ? (in atm)
- <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>
<u><em>Answer: </em></u>
<u><em>The new pressure of the gas is 2.11 atm </em></u>
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For reaction a, the mole number of gas increase from 2 to 4. So the entropy will increase, ΔS>0. For b, the gas will change to solid, so the entropy decrease, ΔS<0.
<h3>
Answer:</h3>
Limiting reagent: Potassium iodide
Mass of the precipitate (PbI₂) is 4.453 g
<h3>
Explanation:</h3>
We are given;
- 60.0 mL of 0.322 M potassium iodide
- 20.0 mL of 0.530 M lead () nitrate
We are required to identify the limiting reactant and determine the mass of the precipitate formed.
<h3>Step 1: Write the balanced equation for the reaction</h3>
- The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;
2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)
<h3>Step 2: Determine the number of moles of the reagents</h3>
Moles of KI
Moles = Molarity × volume
Moles of KI = 0.322 M × 0.060 L
= 0.01932 moles
Moles of KNO₃
Moles = 0.530 M × 0.020 L
= 0.0106 M
From the equation;
- 2 moles of KI reacts with 1 mole of Pb(NO)₂
- Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb(NO₃)₂
- This means, KI is the limiting reagent while Pb(NO₃)₂ is the excess reagent.
<h3>Step 3: Determine the mass of the precipitate PbI₂</h3>
2 moles of KI reacts to produce 1 mole of PbI₂
Therefore;
Moles of PbI₂ = Moles of KI ÷ 2
= 0.01932 moles ÷ 2
= 0.00966 moles
But molar mass of PbI² is 461.01 g/mol
Therefore;
Mass of PbI₂ = 0.00966 moles × 461.01 g/mol
= 4.453 g
Therefore, the mass of the precipitate formed (Pbi₂)is 4.453 g
The answer is 2 more oxygen is needed